Question

Let f(x) = 2x+1 for x <= -2, ax^2+b for -2=1. Find values for a and b such that the function is continuous and use the definition of continuity to justify your answer.

Answer 1

\[f(x)=\begin{cases}2x+1&\text{for }x\le-2\\ax^2+b&\text{for }-2<x<1\\\ln x&\text{for }x\ge1\end{cases}\]
\(f(x)\) is continuous at \(x=-2\) and \(x=1\) when the following are satisfied, respectively:
\[\large\lim_{x\to-2^-}f(x)=\lim_{x\to-2^+}f(x)=f(-2)\\
\large\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)\]

Answer 2

In particular, you want \(a\) and \(b\) that establish the equalities,
\[\large\lim_{x\to-2^-}(2x+1)=\lim_{x\to-2^+}(ax^2+b)=-3\\
\large\lim_{x\to1^-}(ax^2+b)=\lim_{x\to1^+}\ln x=0\]
This gives you two equations:
\[\begin{cases}4a+b=-3\\a+b=0\end{cases}\]